将每行的总数与其余行进行比较

Question

将下面每一行的值与其余行的值进行比较的最佳方法是什么。 假设我想发现 2 行或更多行具有相同的总数（if 语句除外）。

下面的每个索引都源自一个随机函数。

import random main_list = [random.randint(1,9) for iter in range(25)]

row_1_total = main_list[0] + main_list[1] + main_list[2] + main_list[3] + main_list[4]
row_2_total = main_list[5] + main_list[6] + main_list[7] + main_list[8] + main_list[9]
row_3_total = main_list[10] + main_list[11] + main_list[12] + main_list[13] + main_list[14]
row_4_total = main_list[15] + main_list[16] + main_list[17] + main_list[18] + main_list[19]
row_5_total = main_list[20] + main_list[21] + main_list[22] + main_list[23] + main_list[24]

print("Total of row 1 >>>", row_1_total)
print("Total of row 2 >>>", row_2_total)
print("Total of row 3 >>>", row_3_total)
print("Total of row 4 >>>", row_4_total)
print("Total of row 5 >>>", row_5_total)

Answer 1

将行添加到数据框中并使用df.diff().eq(0)或任何比较

import pandas as pd
row_1_total=3
row_2_total=6
row_3_total=7
row_4_total=5
row_5_total=5
df= pd.DataFrame([row_1_total,row_2_total,row_3_total ,row_4_total ,row_5_total  ])
df.diff().eq(0)
print(df.diff().eq(0))
duplicat = df[df.duplicated()]
print(duplicat)

Answer 2

将总和保存在列表中而不是 5 个变量。 如果你失去循环，你可以在每次迭代中获取列表的一部分并将你的代码减少到

main_list = [random.randint(1, 9) for _ in range(25)]
sums = [sum(main_list[i:i+4]) for i in range(0, len(main_list), 5)]
for i in range(len(sums)):
    print(f"Total of row {i + 1} >>>", sums[i])

现在您可以将sums放入set()并比较大小

sums_set = set(sums)
print(len(sums_set) == len(sums))

Answer 3

首先，我建议在稍微不那么笨拙的数据结构中收集总数，例如列表：

row_total = []
row_total.append(main_list[0] + main_list[1] + main_list[2] + main_list[3] + main_list[4])
row_total.append(main_list[5] + main_list[6] + main_list[7] + main_list[8] + main_list[9])
row_total.append(main_list[10] + main_list[11] + main_list[12] + main_list[13] + main_list[14])
row_total.append(main_list[15] + main_list[16] + main_list[17] + main_list[18] + main_list[19])
row_total.append(main_list[20] + main_list[21] + main_list[22] + main_list[23] + main_list[24])
print(row_total[3])

其次，您可以通过像这样在一行中计算总数来避免大量重复（假设main_list在 24 main_list停止）：

row_total = [sum(x[0+n:3+n]) for n in range(0, len(x), 3)]]

你的问题的答案将是这样的：

from collections import Counter
from random import randint

main_list = [randint(0, 10) for _ in range(25)]

row_total = [sum(main_list[0+n:3+n]) for n in range(0, len(main_list), 3)]

duplicate_totals = {t: c for t, c in Counter(row_total).items() if c > 1}

print(main_list)
# a dictionary of totals that show up twice or more often
print(duplicate_totals)

结果将类似于：

[4, 9, 1, 2, 0, 6, 8, 6, 4, 3, 0, 10, 0, 9, 1, 0, 10, 7, 10, 7, 0, 6, 2, 1, 7]
{17: 2}

Answer 4

我想你正在寻找这样的东西：

main_list = [[5,4,6], [1, 2, 3, 4], [4,5,6]]
total = []
for row in main_list:
    total.append(sum(row))


for t in total:
    print("value : ",t, " is at indices : ", total.indices(t))